1

Suppose that I use Finite Difference Method to solve a stable PDE, which is 4th order in $x$ and 2nd order in $t$.

Unfortunately, the discretization in $x$ is irregular ($\Delta x_i$ values are different). Is there a mathematical result for checking the stability of the method (note, $\Delta t$ is uniform, i don't vary it)?

It seems that if I modify the problem to be uniformly discretized, with discretization step $min \Delta x_i$ (so I take the smallest step of the irregular discretization to be the step of the uniform one), then if the modified problem is stable, then the irregular one is stable too.

Is this statement true? Or it is not that simple, and this is dependent on the exact form of the PDE?

geza
  • 293
  • Stability is usually relatively insensitive to the spatial discretization and more strongly sensitive to the time discretization. After all, the time step is usually of the form $x_{n+1}=Ax_n+b$ for some matrix $A$ and vector $b$ (possibly depending on $x$ and/or $n$), and so everything hinges on the spectral properties of $A$. Typically for fine enough meshes, even irregular, the spectral properties of a finite difference operator resemble those of the continuous operator, and so the main issue is about perturbing real-part-zero eigenvalues into real-part-positive eigenvalues... – Ian May 22 '18 at 16:33
  • @Iam, hmm, for the 2nd order case, the CFL condition is $\Delta t/\Delta x<$constant, so it has the same sensitivity, hasn't it? (note, my $\Delta t$ is constant, because I need all the values, I cannot do a big step in time, even if the step would be stable) – geza May 22 '18 at 16:37
  • In effect the CFL condition and similar conditions should really be viewed as "for a fixed discretization, you must take $\Delta t$ this small for stability", i.e. everything really hinges on $\Delta t$. – Ian May 22 '18 at 16:39
  • @Ian, that's okay, I just say that both $\Delta t$ and $\Delta x$ equally important. – geza May 22 '18 at 16:40
  • I can put it a different way: stability comes in three pieces. There is stability of the PDE itself. There is stability of the ODE given after discretizing in space and not time. And there is stability of the difference scheme given after discretizing in space and time. Usually, provided the first one is true in the first place, the second one is not a big deal. The difficulty is passing from the second to the third, which in general requires a time step constraint, which may depend on what your discretization looked like. – Ian May 22 '18 at 16:43
  • (Cont.) But it is easier to do the analysis for the difference scheme if you already know that the high-dimensional ODE that the time discretization is solving is itself stable, as a black box result. – Ian May 22 '18 at 16:43
  • @Ian, my $\Delta t$ is uniform, and the PDE itself is stable (I've edited the question to reflect these). – geza May 22 '18 at 16:46
  • 1
    @Ian, just a simple example: what would be the exact stability condition solving the 2nd order wave equation (centered in time, centered in space approximation), if the discretization is irregular in $x$ (but not in $t$)? Like $x_i$'s are not simply equal to $i\Delta x$, but come from some data? Is it possible to analyze this? – geza May 22 '18 at 16:49
  • I think for the basic wave equation you get a similar requirement to transport, which is that your time step should always be small enough that the traveling waves do not travel more than any one node's distance in a single time unit. Thus you need $\Delta t<\min \Delta x/|c|$. – Ian May 22 '18 at 16:51
  • @Ian, yep, that's what I think too :) And it's logical (because the wave should travel ,,slower'' than the distance of the smallest $\Delta x$), and maybe can be proven. But my case is more complex, as I have a 4th order equation in $x$. That's why I asked my question, is there any rigorous result about this case (which is not just a thought, but a proof)? – geza May 22 '18 at 16:55

1 Answers1

1

just a simple example: what would be the exact stability condition solving the 2nd order wave equation (centered in time, centered in space approximation), if the discretization is irregular in x (but not in t)?

For a standard second-order wave equation in 1D with wave speed $c$, $$\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}-\frac{\partial^2 E}{\partial x^2}=0$$ time-discretized with a second-order accurate finite difference "leapfrog" method, the stability condition on a uniform grid is $$c\Delta_t<\Delta_x$$

On a non-uniform grid it becomes $$c\Delta_t<\frac{2}{\|D\|}$$ where $D$ is the matrix that represents the discretized spatial derivative operator on the non-uniform grid, and $\|D\|=\max_{|x|=1}|Dx|$ is the operator norm.

The proof is not super-complicated. See for example this thesis (eq. 2.113). The proof makes no assumptions on the nature of $D$, so could probably be easily extended to a fourth-order (in space) wave equation.

Wouter
  • 7,673
  • Which operator norm? Or is it really the spectral radius, and so any operator norm suffices? – Ian May 23 '18 at 18:09
  • It is precisely the induced 2-norm, the square root of the first principal component, $|D|=\max_{|x|=1} |Dx|$. This is derived from a spectral radius, not from $\rho(D)$, but from $\rho\left(\left[\begin{matrix}0 & D \ -D^T & 0\end{matrix}\right]\right)$ which equals $|D|$. – Wouter May 24 '18 at 03:54